I will do this part for you \[\int\limits_{0}^{2} dx/(x+1)\] I assume this is what you meant \[\int\limits_{0}^{2} (dx+dx/(x+1)^{2})\]\[let u = (x+1) then du = dx\] let u = (x+1) then du = dx and \[\int\limits\limits dx/(x+1)^{2} =\int\limits\limits du/u ^{2} = u ^{-2+1}/(-2+1)\]=\[-1(1/(x+1)_{0}^{2}=2/3\] Now all you have to do is \[\int\limits_{0}^{2} dx\] Final answer is 8/3.

question guys how do u connect the integral sign to dx \[\int\limits_{?}^{?}\]dx

anonymous

7 years ago

well sjin did u get it? that must be the way u solve it

anonymous

7 years ago

from my calculator the final ans is 2.301987535

anonymous

7 years ago

if you meant \[\int\limits du/(a ^{2} + u ^{2})^{n}\] where a =1 and u = (1/x+1) then you should look this up in a table of integrals

anonymous

7 years ago

if you meant \[\int\limits \sqrt({a}^{2}+u ^{2})du\] where a =1 and u = 1/(x+1) than again the best thing to to do is to look up the integral in a table of integrals or input the integral into a program like Maple. I hope I am not confusing you.